Essential infimum and integral Is the inequality
$$
\int_{r}^{\infty}\operatorname{essinf}\limits_{t<s<\infty}\varphi(s)\frac{dt}{t}\leq \int_{r}^{\infty}\varphi(t)\frac{dt}{t}
$$
true? I think the inequality
$$
\int_{r}^{\infty}\inf_{t\leq s<\infty}\varphi(s)\frac{dt}{t}\leq \int_{r}^{\infty}\varphi(t)\frac{dt}{t}
$$
is true. Since, from the definition of infimum $\varphi(s)\geq \inf\limits_{t\leq s<\infty}\varphi(s)$ for all $s\in[t,\infty)$, in particular for $s=t$. But about the first inequality what should we say? 
 A: If the RHS $\int_r^\infty \varphi(t) \, dt/t$ is infinite, the claim is trivial.
Hence, we may assume $\int_r^\infty \varphi(t) \, dt/t < \infty$. In particular, this implies that $\psi := \varphi \cdot \chi_{(r,\infty)}$ (extended by $0$ to $\Bbb{R}$) is locally integrable on $\Bbb{R}$ (why exactly?).
Hence, almost every point $x \in \Bbb{R}$ is a Lebesgue-point of $\psi$. Now, let $x \in (r,\infty)$ be a Lebesgue point of $\psi$. This implies
$$
|\varphi(x) - \frac{1}{h} \int_x^{x+h} \varphi(y) \, dy| \leq \frac{1}{h} \int_{x}^{x+h} |\varphi(x) - \varphi(y) | \, dy \leq 2 \cdot \frac{1}{2h} \int_{x-h}^{x+h} |\psi(x) - \psi(y)| \, dy \xrightarrow[h\to 0]{} 0,
$$
where in the calculation, $h > 0$ is chosen so small that $(x-h, x+h) \subset (r,\infty)$, so that $\psi(x) = \varphi(x)$ on $(x-h,x+h)$.
But this show
$$
\varphi(x) = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} \varphi(y) \, dy \geq {\rm essinf}_{x  < s < \infty} \varphi(s)
$$
for every Lebesgue-point $x \in (r,\infty)$ of $\psi$ and hence for almost every $x \in (r,\infty)$.
This should allow you to conclude the proof.
